Triangles — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Triangles" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Triangles" — 9 important questions with detailed answers for CBSE board exam prepa…
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NCERT Solutions for Class 10 Mathematics — Triangles. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Triangles — All 9 Questions Solved
Q1. State and explain the Basic Proportionality Theorem (Thales' Theorem). What does it tell us about similar triangles?
Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same ratio.
Statement: In triangle ABC, if DE || BC (DE is parallel to BC), where D is on AB and E is on AC, then:
AD/DB = AE/EC
Explanation:
- The parallel line DE creates two smaller triangles: ADE and ABC
- These triangles are similar (AA similarity — angle A is common, and corresponding angles are equal due to parallel lines)
- Similar triangles have proportional sides
- Therefore, the sides are divided in the same ratio
Application: This theorem helps us find unknown lengths in triangles when parallel lines are present, and it connects to similarity criteria.
Converse: If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
This is a fundamental tool for solving many geometry problems involving proportionality.
Q2. Two triangles ABC and PQR are similar. If AB = 4 cm, BC = 6 cm, CA = 5 cm, and PQ = 6 cm, find QR and RP.
Given: Triangle ABC ~ Triangle PQR
AB = 4 cm, BC = 6 cm, CA = 5 cm
PQ = 6 cm
Find: QR and RP
Step 1: Since ABC ~ PQR, the corresponding sides are proportional.
Corresponding sides are: AB ↔ PQ, BC ↔ QR, CA ↔ RP
Step 2: Write the ratio of similarity.
AB/PQ = BC/QR = CA/RP = k (ratio of similarity)
Step 3: Find the ratio k.
k = AB/PQ = 4/6 = 2/3
Step 4: Find QR.
BC/QR = 2/3
6/QR = 2/3
QR = 6 × 3/2 = 9 cm
Step 5: Find RP.
CA/RP = 2/3
5/RP = 2/3
RP = 5 × 3/2 = 7.5 cm
Answer: QR = 9 cm, RP = 7.5 cm
Q3. In triangle ABC, DE || BC where D is on AB and E is on AC. If AD = 3 cm, DB = 6 cm, and AE = 2 cm, find EC.
Given: Triangle ABC with DE || BC
D is on AB, E is on AC
AD = 3 cm, DB = 6 cm, AE = 2 cm
Find: EC
Step 1: Apply Basic Proportionality Theorem.
Since DE || BC, we have:
AD/DB = AE/EC
Step 2: Substitute the known values.
3/6 = 2/EC
Step 3: Solve for EC.
1/2 = 2/EC
EC = 2 × 2 = 4 cm
Answer: EC = 4 cm
Verification: AD/DB = 3/6 = 1/2 and AE/EC = 2/4 = 1/2 ✓
Q4. Two similar triangles have corresponding sides in the ratio 3:5. If the area of the smaller triangle is 36 cm², find the area of the larger triangle.
Given: Two similar triangles with sides in ratio 3:5
Area of smaller triangle = 36 cm²
Find: Area of larger triangle
Step 1: Recall the relationship between areas of similar triangles.
When two triangles are similar with sides in ratio k:1, their areas are in ratio k²:1.
Step 2: Here the ratio of sides = 3:5
Ratio of areas = (3)²:(5)² = 9:25
Step 3: Let the area of the larger triangle be A.
Area of smaller triangle : Area of larger triangle = 9:25
36:A = 9:25
Step 4: Solve for A.
36/A = 9/25
A = 36 × 25/9
A = 4 × 25
A = 100 cm²
Answer: Area of larger triangle = 100 cm²
Verification: 36:100 = 9:25 ✓
Q5. In triangle ABC, a line parallel to BC intersects AB at D and AC at E such that AD:DB = 2:1. If BC = 15 cm, find DE.
Given: DE || BC in triangle ABC
AD:DB = 2:1, BC = 15 cm
Find: DE
Step 1: Express AD and DB in terms of a variable.
Let AD = 2x and DB = x
Then AB = AD + DB = 2x + x = 3x
Step 2: Find the ratio AD:AB.
AD:AB = 2x:3x = 2:3
Step 3: Triangles ADE and ABC are similar (AA similarity).
Since DE || BC:
∠ADE = ∠ABC (corresponding angles)
∠AED = ∠ACB (corresponding angles)
∠A is common
Step 4: Write the ratio of corresponding sides.
Since triangles ADE and ABC are similar:
AD/AB = DE/BC = AE/AC
Step 5: Find DE.
2/3 = DE/15
DE = 15 × 2/3 = 10 cm
Answer: DE = 10 cm
Q6. In a right triangle ABC with right angle at B, the altitude BD is drawn to the hypotenuse AC. Prove that AB² = AD × AC.
Given: Triangle ABC with ∠B = 90°
BD is the altitude from B to hypotenuse AC
Prove: AB² = AD × AC
Proof:
Step 1: Consider triangles ABD and ABC.
In triangle ABD:
∠ADB = 90° (BD is perpendicular to AC)
∠BAD = ∠BAC (common angle)
Therefore ∠ABD = ∠ABC - 90°
Step 2: In triangle ABC:
∠ABC = 90° (given)
∠BAC = ∠BAC (common)
∠BCA = 90° - ∠BAC
Step 3: Consider triangles ABD and CBA.
In triangle ABD: ∠ADB = 90°, ∠BAD = ∠BAC
In triangle CBA: ∠ABC = 90°, ∠BCA = ∠BCA
Step 4: Show that triangle ABD ~ triangle CAB.
∠ADB = ∠ABC = 90°
∠BAD = ∠BCA (both equal to the same angle in the original triangle)
By AA similarity: triangle ABD ~ triangle CAB
Step 5: Write the ratio of corresponding sides.
AD/AB = AB/AC
Step 6: Cross multiply.
AB × AB = AD × AC
AB² = AD × AC
Proof complete. ✓
Q7. The sides of two similar triangles are in the ratio 4:7. The perimeter of the smaller triangle is 48 cm. Find the perimeter of the larger triangle.
Given: Two similar triangles with sides in ratio 4:7
Perimeter of smaller triangle = 48 cm
Find: Perimeter of larger triangle
Step 1: Recall the property of similar triangles.
For similar triangles, the ratio of perimeters equals the ratio of corresponding sides.
Step 2: If sides are in ratio 4:7, then perimeters are also in ratio 4:7.
Step 3: Let the perimeter of the larger triangle be P.
Perimeter of smaller : Perimeter of larger = 4:7
48:P = 4:7
Step 4: Solve for P.
48/P = 4/7
P = 48 × 7/4
P = 12 × 7
P = 84 cm
Answer: Perimeter of larger triangle = 84 cm
Verification: 48:84 = 4:7 ✓
Q8. Triangle ABC has sides AB = 12 cm, BC = 8 cm, CA = 10 cm. Triangle PQR has sides PQ = 18 cm, QR = 12 cm, RP = 15 cm. Are the triangles similar? Justify your answer.
Given: Triangle ABC with sides AB = 12, BC = 8, CA = 10
Triangle PQR with sides PQ = 18, QR = 12, RP = 15
Check: Are the triangles similar?
Step 1: For triangles to be similar by SSS (Side-Side-Side) similarity criterion, the ratios of corresponding sides must be equal.
Step 2: Arrange the sides in order of magnitude.
Triangle ABC: 8, 10, 12
Triangle PQR: 12, 15, 18
Step 3: Find the ratios of corresponding sides.
Smallest side ratio: 8/12 = 2/3
Middle side ratio: 10/15 = 2/3
Largest side ratio: 12/18 = 2/3
Step 4: Check if all ratios are equal.
8/12 = 10/15 = 12/18 = 2/3
Yes, all corresponding sides are in the same ratio.
Step 5: Apply SSS similarity criterion.
Since the ratios of all corresponding sides are equal, by SSS similarity criterion:
Triangle ABC ~ Triangle PQR
Answer: Yes, the triangles are similar with a ratio of 2:3.
Alternative verification:
8 × 15 = 120 and 12 × 10 = 120 ✓
10 × 18 = 180 and 12 × 15 = 180 ✓
8 × 18 = 144 and 12 × 12 = 144 ✓
Q9. In triangle ABC, points D and E lie on sides AB and AC respectively such that DE || BC. If AB = 10 cm, AD = 4 cm, and AC = 15 cm, find AE and EC.
Given: Triangle ABC with DE || BC
D is on AB, E is on AC
AB = 10 cm, AD = 4 cm, AC = 15 cm
Find: AE and EC
Step 1: Find DB.
DB = AB - AD = 10 - 4 = 6 cm
Step 2: Apply Basic Proportionality Theorem.
Since DE || BC:
AD/DB = AE/EC
4/6 = AE/EC
2/3 = AE/EC
Step 3: Express AE and EC in terms of a variable.
Let AE = 2y and EC = 3y
Then AE + EC = AC
2y + 3y = 15
5y = 15
y = 3 cm
Step 4: Find AE and EC.
AE = 2y = 2 × 3 = 6 cm
EC = 3y = 3 × 3 = 9 cm
Answer: AE = 6 cm, EC = 9 cm
Verification: AD/DB = 4/6 = 2/3 and AE/EC = 6/9 = 2/3 ✓
Also, AE + EC = 6 + 9 = 15 = AC ✓
Triangles — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. In ΔABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = x cm, DB = (x-2) cm, AE = (x+2) cm, and EC = (x-1) cm, find the value of x.
- A. 4 cm
- B. 2 cm
- C. 3 cm
- D. 5 cm
Answer: A. 4 cm — By the Basic Proportionality Theorem (Thales Theorem), if DE || BC, then AD/DB = AE/EC. Substituting the given values: x/(x-2) = (x+2)/(x-1). This leads to x(x-1) = (x-2)(x+2), simplifying to x² - x = x² - 4, which gives x = 4.
MCQ 2. Which of the following is NOT a criterion for the similarity of two triangles?
- A. SAS
- B. AAA
- C. SSS
- D. RHS
Answer: D. RHS — RHS (Right angle-Hypotenuse-Side) is a criterion for congruence of triangles, not for similarity. SAS, AAA, and SSS are the standard criteria for similarity.
MCQ 3. If two triangles are similar, and their areas are 81 cm² and 49 cm² respectively, what is the ratio of their corresponding sides?
- A. 9:7
- B. 7:9
- C. 81:49
- D. 49:81
Answer: A. 9:7 — The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, (side1/side2)² = Area1/Area2 = 81/49. Taking the square root, side1/side2 = √(81/49) = 9/7.
MCQ 4. In ΔPQR, ST || QR, where S is on PQ and T is on PR. If PS = 3 cm, SQ = 6 cm, and PT = 2 cm, what is the length of TR?
- A. 4 cm
- B. 6 cm
- C. 3 cm
- D. 2 cm
Answer: A. 4 cm — By the Basic Proportionality Theorem (BPT), since ST || QR, PS/SQ = PT/TR. Substituting the values: 3/6 = 2/TR. This simplifies to 1/2 = 2/TR, so TR = 4 cm.
MCQ 5. If ΔABC ~ ΔDEF such that AB = 6 cm, BC = 8 cm, AC = 10 cm, and DE = 9 cm, what is the perimeter of ΔDEF?
- A. 36 cm
- B. 24 cm
- C. 30 cm
- D. 40 cm
Answer: A. 36 cm — For similar triangles, the ratio of their corresponding sides is equal to the ratio of their perimeters. The ratio of sides AB/DE = 6/9 = 2/3. The perimeter of ΔABC = 6 + 8 + 10 = 24 cm. So, 24 / Perimeter(ΔDEF) = 2/3, which implies Perimeter(ΔDEF) = 24 * (3/2) = 36 cm.
MCQ 6. In a right-angled triangle ABC, right-angled at B, BD is perpendicular to AC. If AD = 4 cm and CD = 5 cm, what is the length of BD?
- A. 2√5 cm
- B. 3√5 cm
- C. 4√5 cm
- D. 6 cm
Answer: A. 2√5 cm — When an altitude is drawn from the vertex of the right angle to the hypotenuse, the two triangles formed are similar to the original triangle and to each other. Thus, ΔADB ~ ΔBDC. From similarity, AD/BD = BD/CD, which means BD² = AD × CD. So, BD² = 4 × 5 = 20, and BD = √20 = 2√5 cm.
MCQ 7. Two triangles are said to be similar if their corresponding angles are ______ and their corresponding sides are ______.
- A. equal, proportional
- B. proportional, equal
- C. equal, equal
- D. proportional, proportional
Answer: A. equal, proportional — The definition of similar triangles states that their corresponding angles must be equal, and their corresponding sides must be in proportion.
MCQ 8. The ratio of the perimeters of two similar triangles is 4:9. What is the ratio of their areas?
- A. 4:9
- B. 16:81
- C. 2:3
- D. 81:16
Answer: B. 16:81 — If the ratio of perimeters of two similar triangles is a:b, then the ratio of their corresponding sides is also a:b. The ratio of their areas is the square of the ratio of their corresponding sides. So, (4/9)² = 16/81.
MCQ 9. In ΔXYZ, M is a point on XY and N is a point on XZ. If XM = 3 cm, MY = 5 cm, XN = 4.5 cm, and NZ = 7.5 cm, is MN || YZ?
- A. Yes
- B. No
- C. Cannot be determined
- D. Only if XY = XZ
Answer: A. Yes — To check if MN || YZ, we use the converse of the Basic Proportionality Theorem. Calculate the ratios XM/MY = 3/5. And XN/NZ = 4.5/7.5 = 45/75 = 3/5. Since XM/MY = XN/NZ, by the converse of BPT, MN || YZ.
MCQ 10. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This statement is known as:
- A. Basic Proportionality Theorem
- B. Pythagoras Theorem
- C. Thales Theorem
- D. Mid-point Theorem
Answer: B. Pythagoras Theorem — This fundamental relationship in a right-angled triangle is precisely the statement of the Pythagoras Theorem.
MCQ 11. In ΔABC and ΔPQR, ∠A = ∠P and ∠B = ∠Q. Which similarity criterion proves that ΔABC ~ ΔPQR?
- A. SSS
- B. SAS
- C. AAA (or AA)
- D. RHS
Answer: C. AAA (or AA) — If two angles of one triangle are respectively equal to two angles of another triangle, then the third angles must also be equal, and the triangles are similar by the AA (or AAA) similarity criterion.
MCQ 12. ABCD is a trapezium with AB || DC. The diagonals AC and BD intersect each other at point O. If AB = 2CD, what is the ratio of the areas of ΔAOB and ΔCOD?
- A. 4:1
- B. 1:4
- C. 2:1
- D. 1:2
Answer: A. 4:1 — Since AB || DC, ΔAOB ~ ΔCOD by AAA similarity (alternate interior angles and vertically opposite angles). The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. So, Area(ΔAOB) / Area(ΔCOD) = (AB/CD)² = (2CD/CD)² = (2/1)² = 4/1.
MCQ 13. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
- A. 6 m
- B. 5 m
- C. 7 m
- D. 8 m
Answer: A. 6 m — The ladder, the wall, and the ground form a right-angled triangle. By Pythagoras Theorem, (distance from wall)² + (height of window)² = (length of ladder)². So, x² + 8² = 10². x² + 64 = 100, which gives x² = 36, and thus x = 6 m.
MCQ 14. If ΔABC ~ ΔPQR, and AD and PS are medians to BC and QR respectively. If BC = 8 cm and QR = 6 cm, what is the ratio AD/PS?
- A. 4:3
- B. 3:4
- C. 16:9
- D. 9:16
Answer: A. 4:3 — For similar triangles, the ratio of corresponding medians is equal to the ratio of their corresponding sides. Therefore, AD/PS = BC/QR = 8/6 = 4/3.
MCQ 15. A triangle with all three angles equal is called an:
- A. Isosceles triangle
- B. Scalene triangle
- C. Equilateral triangle
- D. Right-angled triangle
Answer: C. Equilateral triangle — An equilateral triangle is defined as a triangle with all three sides equal in length and all three angles equal to 60 degrees.
MCQ 16. The altitude of an equilateral triangle with side 'a' is:
- A. a√3/2
- B. a√2/2
- C. a/2
- D. a√3
Answer: A. a√3/2 — In an equilateral triangle, the altitude bisects the base. Using Pythagoras Theorem, if 'h' is the altitude, h² + (a/2)² = a². So, h² = a² - a²/4 = 3a²/4. Thus, h = a√3/2.
MCQ 17. In ΔABC, right-angled at B, AB = 6 cm and BC = 8 cm. What is the length of AC?
- A. 10 cm
- B. 12 cm
- C. 14 cm
- D. 8 cm
Answer: A. 10 cm — By Pythagoras Theorem in a right-angled triangle, the square of the hypotenuse (AC) is equal to the sum of the squares of the other two sides (AB and BC). So, AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100. Thus, AC = √100 = 10 cm.
MCQ 18. The areas of two similar triangles are 100 cm² and 64 cm². If the median of the first triangle is 10 cm, find the corresponding median of the second triangle.
- A. 8 cm
- B. 6.4 cm
- C. 12.5 cm
- D. 7.5 cm
Answer: A. 8 cm — The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. So, Area1/Area2 = (Median1/Median2)². 100/64 = (10/Median2)². Taking the square root of both sides gives 10/8 = 10/Median2. Therefore, Median2 = 8 cm.
MCQ 19. In ΔABC, D and E are points on AB and AC respectively such that DE || BC. If AD:DB = 3:2 and AC = 15 cm, find AE.
- A. 9 cm
- B. 6 cm
- C. 10 cm
- D. 12 cm
Answer: A. 9 cm — By the Basic Proportionality Theorem, AD/DB = AE/EC. Given AD:DB = 3:2, so AE:EC = 3:2. Let AE = 3k and EC = 2k. Then AC = AE + EC = 3k + 2k = 5k. Since AC = 15 cm, 5k = 15, which means k = 3. Therefore, AE = 3k = 3 * 3 = 9 cm.
MCQ 20. If in two triangles, the corresponding angles are equal, then the triangles are:
- A. Congruent
- B. Similar
- C. Equal in area
- D. Isosceles
Answer: B. Similar — According to the AAA (Angle-Angle-Angle) similarity criterion, if all corresponding angles of two triangles are equal, the triangles are similar.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.